Chapter 22, Problem 77AP

Commercial silver-plating operations frequently use a solution containing the complex $\text{Ag}{\left(\text{CN}\right)}_{\text{2}}{}^{\text{-}}$ ion. Because the formation constant $\left(K_{\text{f}}\right)$ is quite large, this procedure ensures that the free ${\text{Ag}}^{\text{-}}$ concentration in solution is low for uniform electrodeposition. In one process, a chemist added $\text{9}\text{.0 L of 5}\text{.0 }M\text{ NaCN to 90}\text{.0 L of 0}\text{.20 }M{\text{ AgNO}}_{\text{3}}$. Calculate the concentration of free Ag^- ions at equilibrium. See Table 17.5 for $K_{\text{f}}$ value.

Interpretation Introduction

Interpretation:

The concentration of free ${\text{Ag}}^{+}$ ions in a solution at equilibrium is to be calculated with the volume and concentration $\text{NaCN}\text{and}{\text{AgNO}}_{\text{3}}$.

Concept introduction:

When the rate of forward reaction is equal to the backward reaction, a chemical equilibrium is reached.

The general expression of formation constant is represented as follows:

$K_{\text{f}}\text{ = }\frac{\left[{\text{M}}_{\text{x}}{\text{L}}_{\text{y}}\right]}{{\left[\text{M}\right]}^{\text{x}}{\left[\text{L}\right]}^{\text{y}}}$

Here, $\text{M}$ is a metal ion, $\text{L}$ is the ligand, and $x\text{ and }y$ are the coefficients.

The relation between molarity and volume is expressed as:

$M_1{V}_1=MV$

Where $M$ ismolarity and $V$ is volume of the solution.

Answer to Problem 77AP

Solution: $2.2\times 10^{-20}\text{M}$

Explanation of Solution

The reaction is given as follows:

${\text{Ag}}^{+}\left(aq\right)+2{\text{CN}}^{-}\left(aq\right)\rightleftarrows {\text{[Ag}{\left(\text{CN}\right)}_{\text{2}}\text{]}}^{–}\left(aq\right)$.

The expression of $K_f$ for the reaction is given as follows:

$K_{\text{f}}\text{ = }\frac{\left[{{\text{[Ag(CN)}}_{\text{2}}\text{]}}^{-}\right]}{\left[{\text{Ag}}^{\text{+}}\right]{\left[{\text{CN}}^{-}\right]}^{\text{2}}}$

From table 17.5, the value of $K_f$ for the reaction is $\text{1}{\text{.0 × 10}}^{\text{21}}$.

Thus,

$\text{1}{\text{.0 × 10}}^{\text{21}}\text{= }\frac{\left[{{\text{[Ag(CN)}}_{\text{2}}\text{]}}^{-}\right]}{\left[{\text{Ag}}^{\text{+}}\right]{\left[{\text{CN}}^{-}\right]}^{\text{2}}}$ …… (1)

From the above equation, it is clear that the value of $K_{\text{f}}$ is very large. It means that the reaction goes to completion. This assumption allowssolving for the concentration of Ag⁺ at equilibrium.

The initial concentration of ${\text{CN}}^{\text{-}}$ ions is given as follows:

The relation between molarity and volume is expressed as follows:

$M_1{V}_1=MV$

Substitute $\text{5}\text{.0 M}$ for $M_1$, $\text{9}\text{.0}\text{L}$ for $V_1$ and $\text{99}\text{.0}\text{L}$ for $V$ in the above equation as follows:

$\begin{array}{c}M\left(\left[{\text{CN}}^{\text{-}}\right]\right)\text{=}\frac{\text{5}\text{.0 M×}\text{9}\text{.0}\text{L}}{\text{99}\text{.0}\text{L}}\\ \text{=}\text{0}\text{.455}\text{M}\end{array}$

The initial concentration of ${\text{Ag}}^{\text{+}}$ ions is given as follows:

The relation between molarity and volume is expressed as follows:

$M_2{V}_2=MV$

Substitute $\text{0}\text{.20 M}$ for $M_2$, $\text{90}\text{.0}\text{L}$ for $V_2$ and $\text{99}\text{.0}\text{L}$ for $V$ in the above equation as follows:

$\begin{array}{c}M\left(\left[{\text{Ag}}^{\text{+}}\right]\right)=\frac{\left(\text{0}\text{.20 M}\right)\text{×}\left(\text{90}\text{.0}\text{L}\right)}{\text{99}\text{.0}\text{L}}\\ =\text{0}\text{.182}\text{M}\end{array}$

The ICE table for the reaction is given as follows:

$\begin{array}{l} {\text{Ag}}^{+}\left(aq\right)+2{\text{CN}}^{-}\left(aq\right)\rightleftarrows \text{ [Ag}{\left(\text{CN}\right)}_{\text{2}}{\text{]}}^{–}\left(aq\right)\\ \begin{array}{l}Initial\left(\text{M}\right): 0.182 0.455 0\hfill \\ Change\left(\text{M}\right): -0.182 -\left(2\right)\left(0.182\right) +0.182\hfill \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\hfill \\ Equilibrium\left(\text{M}\right): 0 0.0910 0.182\hfill \end{array}\end{array}$

Substitute $\text{0}\text{.182}\text{M}$ for ${{\text{[Ag(CN)}}_{\text{2}}\text{]}}^{-}$ and $\text{0}\text{.0910}\text{M}$ for $\left[{\text{CN}}^{-}\right]$ in equation (1) as follows:

$\begin{array}{c}\text{1}\text{.0}\text{×}{\text{10}}^{\text{21}}\text{=}\frac{\text{0}\text{.182}\text{M}}{\left[{\text{Ag}}^{\text{+}}\right]{\left(\text{0}\text{.0910}\text{M}\right)}^{\text{2}}}\\ \left[{\text{Ag}}^{\text{+}}\right]= 2.2\times {10}^{-20}\text{ M}\end{array}$

Conclusion

The concentration of free ${\text{Ag}}^{+}$ ions in a solution at equilibrium is $2.2\times {10}^{-20}\text{ M}$.